纠错编码习题解答汇总

纠错编码习题解答

第一章

1.1 Solution: p=0.05

(1)The correct decoding P c is P c= P0 =C40p0(1-p)4=0.8145

(2)The decoding error P e is P e = P2+P4 = C42p2(1-p)2 + C44p4(1-p)0 = 0.0135

(3)The decoding failure P f is P f= C41p(1-p)3 + C43p3(1-p) = 0.1720

1.2 Solution:

Because the success rate does not fall below 99%,then the decoding failure P f <1% .And p<<1, P f = P1 = n*0.001*0.999n-1 < 0.01

So n<=10 .then the maximum blocklength n such that the success rate does not fall below 99% is 10.

1.3 Solution: p=0.01

P f = P2 = C42p2(1-p)2 = C42 * 0.012 * 0.992 = 0.000588

So the decoding failure rate is 0.000588.

1.4 Solution:

(a)Error: There is one error

(b)Correct

(c)Failure

(d)Error: There is two error

1.5 Solution:

S1 = v1+v2+v3+v4+v6+v8+v9+v12

S2 = v2+v3+v4+v5+v7+v9+v10+v13

S3 = v3+v4+v5+v6+v8+v10+v11+v14

4 1235781115

1.6 Solution

(1)s=(0000) ~e = 0000 0000 0000 000

~c= ~e+v1 = (000000000000000)+(100010011001001)= (100010011001001)

(2)s=(1011) ~e = 0000 0001 0000 000

~c= ~e+v2 = (000000010000000)+(001001110100110)= (001001100100110)

1.7 Solution

(1) v=(1011 110) s=(110)

~e = (001 0000) ~c=(1001 110)

(2) v=(1100 110) s=(100)

~e = (0000 100) ~c=(1100 010)

(3) v=(0001 011) s=(000)

~e=(0000 000) ~c=(0001 011)

第二章

i j .

2.2 Solution

1 1 1 1 1 1 1 r1-r

2 1 0 0 0 1 0 1

G2= 0 1 1 1 0 1 0 r2-r3 0 1 0 0 1 1 1 = G1

0 0 1 1 1 0 1 0 0 1 0 1 1 0

0 0 0 1 0 1 1 r3-r40 0 0 1 0 1 1

G1 is systematic form.

And every liner coder is equivalent to a systematic linear code

So the (7,4) linear codes generated by G1 and G2 equivalent.

2.3 Solution

(a) C 0 = (000) G = (000000) C 1 = (001) G = (001110) C 2 = (010) G = (010101) C 3 = (011) G = (011011) C 4 = (100) G = (100011) C 5 = (101) G = (101101) C 6 = (110) G = (110110) C 7 = (111) G = (111000)

(b)If p=0 then

If p=1 then

2.4 Solution

Because the (4,3) even-parity code is a linear code , The minimum distance d(C i ,C j )= W min = 2 The error detection limit is L=2-1=1

The error correction is t=(2-1)/2=ly 0.

2.5 Solution

1 1 1 0 1 0 0 0

1 1 1 0 1 0 0 0 0 1 1 1 0 1 0 0 r4-r3-r2-r1 0 1 1 1 0 1 0 0 H= 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 1

So the systematic forms is 1 1 1 0 1 0 0 0

0 1 1 1 0 1 0 0

1 1 0 1 0 0 1 0

1 0 1 1 0 0 0 1

Because H = [P T | I n-k] and G=[I k | P]

Then G = 1 0 0 0 1 0 1 1

0 1 0 0 1 1 1 0

0 0 1 0 1 1 0 1

0 0 0 1 0 1 1 1

2.6 Solution

1 1 1 0 1 0 0 0 1 0 1 1

0 1 1 1 0 1 0 0 H T = 1 1 1 1

H= 1 1 0 1 0 0 1 0 1 1 0 1

1 1 1 1 1 1 1 1 0 1 1 1

1 0 0 1

0 1 0 1

0 0 1 1

0 0 0 1

第三章

3.1 solution

Because x3+1

x4+x+1 x7+x3+1

x7+x4+x3

x4+1

x4+x+1

then q(x)= x3+1 and r(x)= x

check answer : p1(x)q(x)+r(x)= (x3+1)( x4+x+1)+x= x7+x3+1 = p2(x) so the solution is correct.

3.2 Solution